Problem:
Let f be a function with the following properties:
(i) f(1)=1, and
(ii) f(2n)=n⋅f(n) for any positive integer n.
What is the value of f(2100) ?
Answer Choices:
A. 1
B. 299
C. 2100
D. 24950
E. 29999
Solution:
Note that
f(21)=f(2)=f(21)=1f(1)=2020=20f(22)=f(4)=f(22)=2f(2)=2120=21f(23)=f(8)=f(24)=4f(4)=222120=2(1+2)f(24)=f(16)=f(28)=8f(8)=23222120=2(1+2+3)
and in general
f(2n)=2(1+2+3+…+(n−1))=2n(n−1)/2
It follows that f(2100)=2(100)(99)/2=24950.
The problems on this page are the property of the MAA's American Mathematics Competitions